## Boundary conditions and time domain electromagnetic waves

Consider two propagating media: a lossy dielectric medium and a lossless dielectric medium. Thus, the interface that separates them has two tangential components of electric field, one for each medium. One of them, the component of the lossy medium, decays with time, say, ET1 = C * exp (-a * t) * sin (b * t). The other one, by being in a lossless medium, does not decay with time: let's say ET2 = D * sin (d * t), for example.

Classical electromagnetism states that the tangential electric field must be continuous at the interface of separation the two media, ie: C * exp (-a * t) * sin (b * t) = D * sin (d * t) for all t> 0. However, these two equations will never equal for all t due to the factor exp (-a * t), present in Et1 and absent in ET2.

Conclusion: there will never be transmission of electromagnetic energy between two media.

Obviously I am wrong. But where?
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 Recognitions: Science Advisor I wonder if the problem is that the loss term should go as (for a wave propagating in 'z') exp (-az) instead of exp(-at). Then there's no problem. Having a loss term like exp(-at) is a way to introduce resonance in a damped oscillator, but I don't recall seeing that in the context of classical E&M.
 Ok, I will explain the problem a little deeper. In frequency domain, it is usual to analyse a lossless medium by solving maxwell wave equation assuming no sources of electric or magnectic field. This is fine; for a lossless medium. The problem arrive when dealing with lossy media. Consider solving wave equation for a circular cavity which has lossy inner dielectric (finite conductivity). Of course, for a medium with sources, there is no damped vibration in time, since it is possible to design the source in a way it compensates the energy losed due to Joule Effect on inner dielectric. However, let's consider the analysis without sources. In the case of inner lossy medium without sources of electromagnectic field, the wave amplitude MUST decay in time, since the system is losing electric / magnectic energy (again, because of Joule effect). In this case, if we try to solve wave equation in frequency domain, we will find that there are no solutions with real number ressonant frequencies for the circular cavity. It's obvious! Since frequency domain assumes undumped waves (exp{jwt}), it will never solve our problem! In fact, the ressonant frequencies we find by solving the problem in frequency domain are complex numbers; then exp{jwt} = exp{j*(wr + j*wi)*t} = exp{(-wi + j*wr)*t}, that is, a dumped wave!! Therefore, the certain way to solve this problem is using time domain Maxwell wave equations. In time domain, the factor exp{-at} appears. Furthermore, if we have two media, one of them being lossy (finite conductivity) and the other one being lossless, there is no way to apply tangential electric (or tangential magnectic) bondary conditions (1st post of this topic). Why?

Recognitions:

## Boundary conditions and time domain electromagnetic waves

I'm a little confused by your post. As best I know, the wave equation for a lossy medium is something like:

$$\nabla \times \nabla \times E = -\mu \frac{\partial J}{\partial t} - \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}}$$

And then substituting Ohm's law J = $\sigma$ E :

$$\nabla \times \nabla \times E = -\mu \sigma \frac{\partial E}{\partial t} - \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}}$$

I'd have to look up the solution in a book- but I know there is an analytical solution. Matching this solution to the source-free solution:

$$\nabla \times \nabla \times E = - \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}}$$

at a boundary should be straightforward, have you tried?

 actually, equation is: $$\nabla^{2} E = \mu \sigma \frac{\partial E}{\partial t} + \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}}$$ Suppose there is only $E_{x}$, so: $E = \hat{a}_{x} E_{x}$ and equation can be rewriten as $$\nabla^{2} E_x = \mu \sigma \frac{\partial E_x}{\partial t} + \mu\epsilon \frac{\partial^{2} E_x}{\partial t ^{2}}$$ This equation is separable in cartesian, cylindrical, spherical and some other coordinate systems. If we choose $$E_x = f(x)g(y)h(z)T(t)$$ and rewrite this equantion, we will find that: $$\mu \sigma \dot{T} + \mu \epsilon \ddot{T} = -\tau^{2} T \quad \tau \in \mathbb{R}$$ The solution is a damped wave on time. The same procedure can be applied for the lossless medium, giving a sinusoidal wave for T(t). Now suppose the boundary is at z = 0. Therefore, the tangential component of the electric field is $$E_x$$ and boundary conditions state that, for both media, they should equal. However, T1 will never be equal to T2 for all T, because T1 is dumped and T2 is not.

Recognitions:
 Quote by ashade actually, equation is: $$\nabla^{2} E = \mu \sigma \frac{\partial E}{\partial t} + \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}}$$
Hang on- $\nabla \times \nabla \times E = \nabla(\nabla \bullet E) - \nabla^{2} E$. This can be reduced to $\nabla \times \nabla \times E = \nabla\rho - \nabla^{2} E$.

Since we are allowing mobile charges, I can see that the charge density gradient will not be zero, and could allow a discontinuity in the slope of E.

 Quote by Andy Resnick Hang on- $\nabla \times \nabla \times E = \nabla(\nabla \bullet E) - \nabla^{2} E$. This can be reduced to $\nabla \times \nabla \times E = \nabla\rho - \nabla^{2} E$. Since we are allowing mobile charges, I can see that the charge density gradient will not be zero, and could allow a discontinuity in the slope of E.
You are missing the part that I am assuming a source free media, hence no mobile charges allowed.

 Quote by ashade actually, equation is: $$\nabla^{2} E = \mu \sigma \frac{\partial E}{\partial t} + \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}}$$ Suppose there is only $E_{x}$, so: $E = \hat{a}_{x} E_{x}$ and equation can be rewriten as $$\nabla^{2} E_x = \mu \sigma \frac{\partial E_x}{\partial t} + \mu\epsilon \frac{\partial^{2} E_x}{\partial t ^{2}}$$

The solution can still be written as
$$E(r,t)=E_0 e^{i(kr-\omega t)}$$
with $$k^2=\mu\epsilon\omega^2 + i \mu\sigma\omega$$

Whoose Im part is
$$Im{k}=\omega\sqrt{\epsilon\mu}(\sqrt{1+\sigma^2/\epsilon^2\omega^2}-1) /2$$
responsible for the damping inside the metal.

 Quote by |squeezed> The solution can still be written as $$E(r,t)=E_0 e^{i(kr-\omega t)}$$ with $$k^2=\mu\epsilon\omega^2 + i \mu\sigma\omega$$ Whoose Im part is $$Im{k}=\omega\sqrt{\epsilon\mu}(\sqrt{1+\sigma^2/\epsilon^2\omega^2}-1) /2$$ responsible for the damping inside the metal.
Sure it can, but it doesn't help answering my question. Diving a little deeper in book "Advanced Engineering Electromgnetics" - Constantine A. Balanis, I ran into the continuity equation, which states:

$$\nabla \cdot (J_i + \sigma E) = -\dot{q_{ev}}$$. Therefore, since $\sigma$ is not null, there must be a charge density in the boundary separanting the 2 media, hence, allowing the discontinuity. Andy Resnick was right.
 I thought the problem was to get the wave's decay correctly through k (and not ω). What exactly is / was your question ?
 I was wrong, the problem is yet not solved. The question is about boundary conditions in the interface of separation of 2 media: one is lossy and the other is lossless. In the lossless media, the wave is non-decaying. In the lossy media, the wave is decaying in time. Since I am assuming no magnectic current sources, $E_{1t} = E_{2t}$, that is, the tangential electric field at the boundary MUST be continuous. However, since the medium 1 is lossy, electric field $E_{1t}$ is decaying with time and, therefore, will never equal $E_{2t}$ (not decaying with time). This is wierd! And I just don't know where I am wrong.

Recognitions:
 Quote by ashade You are missing the part that I am assuming a source free media, hence no mobile charges allowed.
Then why are you keeping the current J?

 Quote by Andy Resnick Then why are you keeping the current J?
I'm keeping a conduction current, not a source current. There's a difference. Anyways, you are wrong: The tangential electric field is discontinuous if, and only if, there is an impressed magnectic current, which is not the case. It has nothing to do with electric charge density. Actually, boundary conditions state that: $n \times (E_1 - E_2) = M_i$. Therefore, nothing to do with free electric charges.

I still think the classical model can do with this. Actually, a discontinuity in conductivity is not physcally realizable. What happens in reality is that, when we put 2 materials together, the boundary contains a mix of molecules of both media that interact together to create a smooth (continuous and differentiable) transition of medium conductivity in space.

Recognitions:
 Quote by ashade I'm keeping a conduction current, not a source current. There's a difference. Anyways, you are wrong: The tangential electric field is discontinuous if, and only if, there is an impressed magnectic current, which is not the case. It has nothing to do with electric charge density. Actually, boundary conditions state that: $n \times (E_1 - E_2) = M_i$. Therefore, nothing to do with free electric charges. I still think the classical model can do with this. Actually, a discontinuity in conductivity is not physcally realizable. What happens in reality is that, when we put 2 materials together, the boundary contains a mix of molecules of both media that interact together to create a smooth (continuous and differentiable) transition of medium conductivity in space.
You really lost me here.... in any case, my goal was to help you resolve your conceptual difficulty, so I'm glad that happened.
 Recognitions: Gold Member Science Advisor And why wouldn't the field in the lossless medium not be decaying with time? If the incident field is coming from a lossy medium that is decaying in time, then obviously the transmitted field would also be decaying in time since the incident field is gradually losing its amplitude.

 Quote by Born2bwire And why wouldn't the field in the lossless medium not be decaying with time? If the incident field is coming from a lossy medium that is decaying in time, then obviously the transmitted field would also be decaying in time since the incident field is gradually losing its amplitude.
You have a point. However, since I'm assuming no sources, it has no meaning in saying "the incident field is coming from a lossy medium". Moreover, a decaying wave cannot solve Maxwell equation $\nabla^2 E_x = \mu \epsilon \partial^2 {E}_x / \partial t^2$ - the equation for the lossless media.

 Quote by Andy Resnick You really lost me here.... in any case, my goal was to help you resolve your conceptual difficulty, so I'm glad that happened.